Comparing the derivation of the Deterministic Policy Gradient Theorem with the standard Policy Gradient Theorem
In the policy gradient theorem, we don't need to write $r$ as a function of $a$ because the only time we explicitly 'see' $r$ is when we are taking the expectation with respect to the policy. For the first couple lines of the PG theorem we have
\begin{align}
\nabla v_\pi(s) &amp;= \nabla \left[ \sum_a \pi(a|s) q_\pi (s,a) \right] \;, \\ &amp;= \sum_a \left[ \nabla \pi(a|s) q_\pi(s,a) + \pi(a|s) \nabla\sum_{s',r} p(s',r|s,a)(r+ v_\pi(s')) \right] \; ;
\end{align}
you can see that we are taking expectation of $r$ with respect to the policy, so we don't need to write something like $r(s,\pi(a|s))$ (especially because this notation doesn't really make sense for a stochastic policy). This is why we don't need to take the derivative of $r$ with respect to the policy parameters. Now, the next line of the PG theorem is
$$\nabla v_\pi(s) = \sum_a \left[ \nabla \pi(a|s) q_\pi(s,a) + \pi(a|s)\sum_{s'} p(s'|s,a) \nabla v_\pi(s') \right] \; ;$$
so now we have an equation similar to the bellman equation in terms of the $\nabla v_\pi(s)$'s, so we can unroll this repeatedly meaning we never have to take an explicit derivative of the value function.
For the deterministic gradient, this is a bit different. In general we have
$$v_\pi(s) = \mathbb{E}_\pi[Q(s,a)] = \sum_a \pi(a|s) Q(s,a)\;,$$
so for a deterministic policy (denoted by $\pi(s)$ which represents the action taken in state $s$) this becomes
$$v_\pi(s) = Q(s,\pi(s))$$
because the deterministic policy has 0 probability for all actions except one, where it has probability one.
Now, in the deterministic policy gradient theorem we can write
$$\nabla v_\pi(s) = \nabla Q(s,\pi(s)) = \nabla \left(r(s, \pi(s)) + \sum_{s'} p(s'|s,a)v(s') \right)\;.$$
We have to write $r$ explicitly as a function of $s,a$ now because we are not taking an expectation with respect to the actions because we have a deterministic policy. Now, if you replace where I have written $\nabla$ with the notation you have used for the derivatives you will arrive at the same result and you'll see why you need to use the chain rule, which I believe you understand because your question was more why don't we use the chain rule for the normal policy gradient, which I have hopefully explained -- it is essentially because of how an expectation over the action space works with a deterministic policy vs. a stochastic policy.
Another way to think of this is as follows -- the term you're concerned with is obtained by expanding $\nabla q_\pi(s,a) = \nabla \sum_{s', r}p(s',r|s,a)(r(s,a) + v_\pi(s'))$. Because, by definition of the $Q$ function, we have conditioned on knowing $a,s$ then $a$ is completely independent of the policy in this scenario - we could even condition on an action that the policy would have 0 probability for - thus the derivative of $r$ with respect to the policy parameters is 0.
However, in the deterministic policy gradient we are taking $\nabla q_\pi(s, \pi(s)) = \nabla \left(r(s, \pi(s)) + \sum_{s'} p(s'|s,a) v_\pi(s')\right)$ -- here $r$ clearly depends on the policy parameters because the action taken was the deterministic action given by the policy in the state $s$, thus the derivative wrt the policy parameters is not necessarily 0!